With your help, this is the proof that I have come up with if anyone reading this at a later time is interested.
My Solution :
First, suppose that $m(\mathbb R) = n(\mathbb R) < \infty$. We show that $m=n$ on $\mathcal B$. Consider the collection $\mathcal G = \{B \in \mathcal B: m(B) = n(B)\}$. We observe that $\mathcal G$ is a monotone class:
Suppose $A_i \uparrow A$ where $A_i \in \mathcal G$ for all $i \in \mathbb N$ and define $B_1 = A_1$, $B_2 = A_2 - A_1$, $\dots$, $B_n = A_n - \cup_{i=1}^{n-1}A_i$ so that $\cup_{i=1}^\infty A_i = \cup_{i=1}^\infty B_i$ where $\{B_i\}$ are pairwise disjoint. Notice that for $C,D \in \mathcal G$ with $C \subset D$, we have
$$m(D-C) = m(D) - m(C) = n(D) - n(C) = n(D-C),$$
since $m$ and $n$ are finite measures. So $B_i \in \mathcal G$ for all $i \in \mathbb N$. Then $A \in \mathcal G$, since $$m(A) = m(\cup_{i=1}^\infty B_i) = \sum_{i=1}^\infty m(B_i) = \sum_{i=1}^\infty n(B_i) = n(\cup_{i=1}^\infty B_i) = n(A)$$
Now suppose that $A_i \downarrow A$ where $A_i \in \mathcal G$ for all $i \in \mathbb N$. Then,
$$m(A) = \lim_{n\to\infty}m(A_n) = \lim_{n\to\infty}n(A_n) = n(A)$$
since $m$ and $n$ are finite measures. So $A \in \mathcal G$, which makes $\mathcal G$ a monotone class. More specifically, it makes $\mathcal G$ a monotone class that contains $\mathcal C = \{(a,b):a,b\in\mathbb R\} \cup \{\mathbb R,\emptyset\}$. Thus, $\mathcal M(\mathcal C) \subseteq \mathcal G$ where $\mathcal M(\mathcal C)$ represents the smallest monotone class containing $\mathcal C$.
Since $\mathcal C$ is a collection of subsets of $\mathbb R$ closed under finite intersections (any intersection is either an open interval or the emptyset) containing $\mathbb R$, we can apply the Monotone Class Theorem. Thus, we get that $\mathcal M(\mathcal C) = \sigma(\mathcal C) = \mathcal B$, and more importantly, $\mathcal B = \mathcal M(\mathcal C) \subseteq \mathcal G$. By definition $\mathcal G \subseteq \mathcal B$, so we have $\mathcal G = \mathcal B$ and $m = n$ on $\mathcal B$.
Now suppose that one of the measures is not finite and consider the following increasing sequence of sets $A_1 = (-1,1)$, $A_2 = (-2,2)$, $\dots$, $A_k = (-k,k)$. Define $m_k(B) = m(B\cap A_k)$ and $n_k(B) = n(B \cap A_k)$ for all $B \in \mathcal B$. Then, $m_k, n_k$ are measures for every $k \in \mathbb N$, and moreover, are finite by monotonicity of measures (they are subsets of sets with finite measure). Therefore, since $m_k(\mathbb R) = n_k(\mathbb R) < \infty$ for all, we can apply the finite case as proved above to get that $m_k = n_k$ on $\mathcal B$ for all $k \in \mathbb N$. Then for any $B \in \mathcal B$, we have
$$m(B) = \lim_{k\to\infty}m(B \cap (-k,k)) = \lim_{k\to\infty}m_k(B) = \lim_{k\to\infty}n_k(B) = \lim_{k\to\infty}n(B\cap (-k,k)) = n(B).$$
Thus, $m=n$ on $\mathcal B$.
